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Ow long does it take for the charge to build up to 9.0 µC? RC circuits

  1. Oct 26, 2005 #1
    Hello everyone, I almost got all the parts to this problem:
    In an RC series circuit, = 11.0 V, R = 1.10 M, and C = 1.80 µF.

    (a) Calculate the time constant.
    1.98 s
    (b) Find the maximum charge that will appear on the capacitor during charging.
    19.8 µC
    (c) How long does it take for the charge to build up to 9.0 µC?
    s

    I tried the following:
    q(t) = Q[1-e^(-t/RC)]
    q(t) = 9.0e-6[1-e^(-1.98)] = 7.75e-6s, which is wrong;
    Since i know t = RC, and i found RC = 1.98, would it be
    q(t) = 9.0e-6[1-e^(-1)] ?

    I just tried this and its also wrong.
     
  2. jcsd
  3. Oct 26, 2005 #2

    daniel_i_l

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    Gold Member

    In your equation q(t) should equal 9.0mc since that is the charge that has built up after time t. and Q which is the total charge should be 19.8mc, then just solve for time.
     
  4. Oct 26, 2005 #3
    thanks! i'm stuck on how i'm suppose to solve for t...
    i have -(19.8E-6)(e^(-t/1.98));
    do i take the natural log of both sides or somthing? but u can't take the natural log of a negative number!
     
  5. Oct 27, 2005 #4

    daniel_i_l

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    Gold Member

    you know that:
    9.0E-6 = 19.8E-6[1-e^(-t/1.98)] = 19.8E-6 - 19.8E-6*e^(-t/1.98) so:

    -10.8E-6 = -19.8E-6*e^(-t/1.98) /-1
    10.8E-6 = 19.8E-6*e^(-t/1.98)
    now you can take the natural log of both sides
     
  6. Oct 27, 2005 #5
    Hey thanks for the responce!
    I took the ln of both sides and got:
    -11.44 = (-t/1.98)*ln(19.8E-6);
    I end up with a negative number of -10.8, did I do this wrong? Sorry its been awhile since I did stuff with ln
     
  7. Oct 27, 2005 #6
    you didn't take the natural logarithm of each side in the correct manner.
    First divide through by 19.8*10^-6 to get rid of that term on the RHS of the expression.

    So you would then have:

    0.54 = e^(-t/1.98)

    Take the natural log of that and you have:

    ln(0.54) = -t/1.98
    This will give you a positive number
     
  8. Oct 27, 2005 #7
    Thanks!! But i submitted that as my answer: t = .311 and it was also incorrect!! hm...
     
  9. Oct 27, 2005 #8
    ohh my bad hah thanks big man, it worked great!!!
     
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